Pump Fan Efficiency

Overview

Engineering reference data for Pump Fan Efficiency in pumps.

Key Formulas

Pump Power

$P = \frac{Q \cdot H \cdot \rho \cdot g}{\eta}$

Hydraulic power / efficiency.

NPSH Available

$NPSH_a = \frac{P_s}{\rho g} + \frac{v_s^2}{2g} - \frac{P_v}{\rho g}$

Net Positive Suction Head available.

Affinity Laws

$Q \propto n, \quad H \propto n^2, \quad P \propto n^3$

Flow, head, power vs speed.

Variables

Types of Efficiency Losses

Pump and fan efficiency is affected by three distinct types of losses that combine to determine overall performance.

Hydraulic Losses

Hydraulic losses arise from the pump or fan construction, caused by:

• Friction between the fluid and internal walls
• Acceleration and retardation of the fluid
• Changes in fluid flow direction

$\eta_h = \frac{w}{w + w_l}$

where $w$ is the specific work from the pump or fan (J/kg) and $w_l$ is the specific work lost due to hydraulic effects (J/kg).

Mechanical Losses

Mechanical components such as transmission gears and bearings create losses that reduce power transfer from the motor shaft to the impeller.

$\eta_m = \frac{P - P_l}{P}$

where $P$ is the power transferred from the motor to the shaft (W) and $P_l$ is the power lost in the transmission (W).

Volumetric Losses

Fluid leakage between the back surface of the impeller hub plate and the casing, or through other components, causes volumetric losses.

$\eta_v = \frac{q}{q + q_l}$

where $q$ is the volume flow out of the pump or fan (m³/s) and $q_l$ is the leakage volume flow (m³/s).

Worked Example: Hydraulic Efficiency

An inline water pump operates between 1 bar and 10 bar with water density $\rho = 1000 \text{ kg/m}^3$ and hydraulic efficiency $\eta_h = 0.91$.

Actual water head:

$h = \frac{p_2 - p_1}{\rho g} = \frac{(10 \times 10^5) - (1 \times 10^5)}{1000 \times 9.81} = 91.7 \text{ m}$

Specific work required:

$w_c = \frac{g \cdot h}{\eta_h} = \frac{9.81 \times 91.7}{0.91} = 988.6 \text{ J/kg}$

Design head:

$h_{design} = \frac{w_c}{g} = \frac{988.6}{9.81} = 100.8 \text{ m}$

Practical Considerations

The losses in a pump or fan convert to heat transferred to the fluid and surroundings. As a rule of thumb, the temperature increase in a fan transporting air is approximately 1°C.

References

• Original Source